Prove a geometric inequality, if 3 numbers are satisfying the condition I just got this problem, but I have no idea on how to prove that.
Prove that if $x,y,z\in\mathbb{R},\ x,y,z\ge 0$ and $2\cdot(x\cdot z+x\cdot y+y\cdot z)+3\cdot x\cdot y\cdot z = 9$, then $(\sqrt x + \sqrt y + \sqrt z )^4 \ge 72$. This is a geometric inequality.
Can anyone help me, please? Any kind of help (solutions, hints etc) is really appreciated. Thank you!
NOTE: I REALLY DON'T KNOW WHAT TITLE SHOULD I WRITE FOR THIS POST, SO PLEASE LEAVE A COMMENT IF YOU HAVE AN IDEA FOR THE POST TITLE. THANK YOU.
 A: Here is an attempt (I think in the good direction, thus a hint, as you wish it) to solve the problem, by putting it in a more tractable form with the elementary symmetrical polynomials in 3 variables, thanks to a change of variables.
Again, let us make it clear : it is not a solution (but I still work on it !).
Let us write the condition and the targetted constraint in order that the text is self-contained:
$$2(xz+xy+yz)+3xyz=9 \ \ \ (1)$$
and 
$$(\sqrt x + \sqrt y + \sqrt z )^4 \ge 72 \ \ \ (2)$$
or by considering power 4 as two successive squarings:
$$(x+y+z+2(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}))^2 \geq 72$$
Let us make the change of variables : 
$$a=\sqrt{yz}, \ \ b=\sqrt{xz}, \ \ c=\sqrt{xy} \ \ (3)$$
the reciprocal formulas being (due to the assumed strict positivity of $x,y,z$):
$$x=\dfrac{bc}{a}, y=\dfrac{ca}{b}, z=\dfrac{ab}{c} \ \ \ (4) $$
Formula (1) becomes
$$2(a^2+b^2+c^2)+3abc=9 \ \ \ (1')$$
or 
$$2(a+b+c)^2-4(ab+bc+ca)+3abc=9 \ \ \ (1'')$$
and constraint to be established becomes:
$$((x+y+z)+2(a+b+c))^2 \geq 72 \ \ \ (2')$$
But, using (4),
$$x+y+z=\dfrac{(bc)^2+(ca)^2+(ab)^2}{abc}=\dfrac{(bc+ca+ab)^2-2abc(a+b+c)}{abc} \ \ \ (5)$$
Plugging (5) in (3'), one gets, for the constraint to be established:
$$\left(\dfrac{(bc+ca+ab)^2}{abc}\right)^2 \geq 72 \ \ \ (2'')$$
Thus, setting 
$$\begin{cases}s_1&=&a+b+c\\s_2&=&ab+ac+bc\\s_3&=&abc\\\end{cases}$$
Thus, the simplified problem is as follows :
Show that, under the condition (1'')
$$2s_1^2-4s_2+3s_3=9$$
(and knowing that all these quantities are positive) we are due to have (2''), i.e.,
$$s_2^4 \leq 72 s_3^2$$
Here, I am a little stucked... Newton identities (https://en.wikipedia.org/wiki/Newton%27s_identities) might may be of some help ?
A: We need to prove that $x+y+z\geq\sqrt[4]{72}$, where $x$, $y$ and $z$ are non-negatives such that
$$\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=9$$
Let $x+y+z<\sqrt[4]{72}$, $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=\sqrt[4]{72}$.
Hence, $k<1$ and $9=\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=k^4\sum\limits_{cyc}(2a^2b^2+k^2a^2b^2c^2)<$
$<\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)$, which is a contradiction because we'll prove now that 
$$\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)\leq9$$
Indeed, we need to prove that
$$9\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^6\geq2(a^2b^2+a^2c^2+b^2c^2)\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^2+3a^2b^2c^2$$ or
$$(a+b+c)^6\geq16(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+144\sqrt2a^2b^2c^2$$
Since by AM-GM 
$(a+b+c)^4=(a^2+b^2+c^2+2(ab+ac+bc))^2\geq$
$\geq\left(2\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\right)^2=8(a^2+b^2+c^2)(ab+ac+bc)$,
it remains to prove that
$$(a+b+c)^2(a^2+b^2+c^2)(ab+ac+bc)\geq2(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+18\sqrt2a^2b^2c^2$$ or
$$(a+b+c)^2((a^2+b^2+c^2)(ab+ac+bc)-2(a^2b^2+a^2c^2+b^2c^2))\geq18\sqrt2a^2b^2c^2$$ or
$$(a+b+c)^2\sum\limits_{cyc}(a^3b+a^3c-2a^2b^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$  or
$$(a+b+c)^2\sum\limits_{cyc}(ab(a-b)^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$ 
Thus, it remains to prove that $(a+b+c)^3\geq18\sqrt2abc$, which is AM-GM:
$(a+b+c)^3\geq27abc\geq18\sqrt2abc$. Done!
